The following table presents the average price in dollars for a dozen eggs and a gallon of milk for each month from February through November.

Dozen Eggs
1.55
1.75
1.63
1.62
1.50
1.37
1.50
1.63
1.83
1.77

Gallon of Milk
3.07
3.08
3.07
3.14
3.26
3.43
3.74
3.81
3.84
3.84

Compute the least-squares regression line for predicting the price of milk from the price of eggs. Round the slope and y-intercept to at least four decimal places.

7 answers

To compute the least-squares regression line, we will use the data given for the average prices of a dozen eggs and a gallon of milk. The least-squares regression line is given in the form:

\[ y = mx + b \]

where \( y \) is the dependent variable (price of milk), \( x \) is the independent variable (price of eggs), \( m \) is the slope of the line, and \( b \) is the y-intercept.

Let’s denote:

  • \( x_i \) = average price of eggs in month \( i \)
  • \( y_i \) = average price of milk in month \( i \)

We are given the following data:

\[ \begin{aligned} \text{Dozen Eggs} & : 1.55, 1.75, 1.63, 1.62, 1.50, 1.37, 1.50, 1.63, 1.83, 1.77 \ \text{Gallon of Milk} & : 3.07, 3.08, 3.07, 3.14, 3.26, 3.43, 3.74, 3.81, 3.84, 3.84 \end{aligned} \]

The formulae needed to calculate \( m \) (slope) and \( b \) (y-intercept) are as follows:

  1. Calculate \( n \), the number of data points.

  2. Compute sums:

    • \( \sum x \)
    • \( \sum y \)
    • \( \sum xy \)
    • \( \sum x^2 \)
    • \( \sum y^2 \)
  3. Calculate the slope \( m \) using:

\[ m = \frac{n(\sum xy) - (\sum x)(\sum y)}{n(\sum x^2) - (\sum x)^2} \]

  1. Calculate the y-intercept \( b \) using:

\[ b = \frac{\sum y - m(\sum x)}{n} \]

Step-by-step calculations:

  1. Count the data points:

    • \( n = 10 \)
  2. Calculate the sums:

Let's calculate each required sum:

\[ \begin{aligned} \sum x & = 1.55 + 1.75 + 1.63 + 1.62 + 1.50 + 1.37 + 1.50 + 1.63 + 1.83 + 1.77 = 16.10 \ \sum y & = 3.07 + 3.08 + 3.07 + 3.14 + 3.26 + 3.43 + 3.74 + 3.81 + 3.84 + 3.84 = 31.40 \ \sum xy & = (1.55 \cdot 3.07) + (1.75 \cdot 3.08) + (1.63 \cdot 3.07) + (1.62 \cdot 3.14) + (1.50 \cdot 3.26) + (1.37 \cdot 3.43) + (1.50 \cdot 3.74) + (1.63 \cdot 3.81) + (1.83 \cdot 3.84) + (1.77 \cdot 3.84) \ & = 4.7765 + 5.3850 + 4.9991 + 5.0808 + 4.8900 + 4.7031 + 5.6100 + 6.1973 + 7.0368 + 6.7976 \ & = 56.4700\ \sum x^2 & = (1.55^2) + (1.75^2) + (1.63^2) + (1.62^2) + (1.50^2) + (1.37^2) + (1.50^2) + (1.63^2) + (1.83^2) + (1.77^2) \ & = 2.4025 + 3.0625 + 2.6569 + 2.6244 + 2.25 + 1.8769 + 2.25 + 2.6569 + 3.3489 + 3.1329 \ & = 24.0569 \end{aligned} \]

  1. Calculate the slope \( m \):

\[ \begin{aligned} m & = \frac{10(56.4700) - (16.10)(31.40)}{10(24.0569) - (16.10)^2} \ & = \frac{564.70 - 505.34}{240.569 - 259.21} \ & = \frac{59.36}{-18.641} \ & \approx -3.1852 \end{aligned} \]

  1. Calculate the y-intercept \( b \):

\[ \begin{aligned} b & = \frac{31.40 - (-3.1852)(16.10)}{10} \ & = \frac{31.40 + 51.34}{10} \ & = \frac{82.74}{10} \ & = 8.2740 \end{aligned} \]

Final least-squares regression line:

The least-squares regression line to predict the price of milk from the price of eggs is:

\[ y = -3.1852x + 8.2740 \]

So, the rounded values are:

  • Slope \( m \approx -3.1852 \)
  • Y-intercept \( b \approx 8.2740 \)
If the price of eggs differs by $0.35 from one month to the next, by how much would you expect the price of milk to differ? Round your answer to two decimal places.

To find the expected difference in the price of milk based on a $0.35 difference in the price of eggs, we use the slope \( m \) from the least-squares regression line that we calculated earlier.

Recall that the slope \( m \) is approximately \(-3.1852\). This slope indicates that for each unit increase in the price of eggs (here $1.00), the price of milk is expected to decrease by about $3.1852.

Since the price of eggs is differing by $0.35, we can calculate the expected difference in the price of milk (\(\Delta y\)) using the slope:

\[ \Delta y = m \cdot \Delta x \]

where

  • \( \Delta x = 0.35 \) (the difference in the price of eggs),
  • \( m = -3.1852 \).

Substituting the values into the equation:

\[ \Delta y = -3.1852 \cdot 0.35 \]

Calculating this gives:

\[ \Delta y \approx -1.11382 \]

Rounding to two decimal places, we find:

\[ \Delta y \approx -1.11 \]

Thus, if the price of eggs differs by $0.35 from one month to the next, you would expect the price of milk to differ by approximately $1.11.

your math is wrong. y=2.4337 + 0.6156x

now compute the difference

Apologies for the confusion in the previous calculations. Let's use the regression equation you provided:

Given the equation: \[ y = 2.4337 + 0.6156x \]

Here:

  • \(y\) = price of milk
  • \(x\) = price of eggs

To find how much the price of milk would change with a difference in the price of eggs of $0.35, we can simply multiply the slope (\(0.6156\)) by the change in \(x\) (\(0.35\)):

Calculation:

\[ \Delta y = 0.6156 \cdot \Delta x \]

Substituting \(\Delta x = 0.35\):

\[ \Delta y = 0.6156 \cdot 0.35 \]

Calculating this gives:

\[ \Delta y = 0.215460 \]

Rounding to two decimal places:

\[ \Delta y \approx 0.22 \]

Thus, if the price of eggs differs by $0.35 from one month to the next, you would expect the price of milk to differ by approximately $0.22. Thank you for your patience!

Predict the price of milk in a month when the price of eggs is 1.96. Round the answer to two decimal places.

To predict the price of milk when the price of eggs is \(1.96\), we can use the regression equation you provided:

\[ y = 2.4337 + 0.6156x \]

where \(y\) is the predicted price of milk and \(x\) is the price of eggs.

Now, substituting \(x = 1.96\) into the equation:

\[ y = 2.4337 + 0.6156(1.96) \]

Calculating the product:

\[ 0.6156 \cdot 1.96 \approx 1.208776 \]

Now, adding this to the constant term:

\[ y \approx 2.4337 + 1.208776 \approx 3.642476 \]

Rounding to two decimal places:

\[ y \approx 3.64 \]

Thus, the predicted price of milk when the price of eggs is $1.96 is approximately $3.64.